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stira [4]
2 years ago
7

How many Joules does it take to raise 1 g of water 2°C

Chemistry
1 answer:
Romashka-Z-Leto [24]2 years ago
4 0

Answer:

Quantitative experiments show that 4.18 Joules of heat energy are required to raise the temperature of 1g of water by 1°C. Thus, a liter (1000g) of water that increased from 24 to 25°C has absorbed 4.18 J/g°C x 1000g x 1°C or 4180 Joules of energy.

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I think the answer is b because that is true 
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Whats does x stand for?
Furkat [3]

x is the chemical symbol of the element and it must correspond to the atomic number

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2 years ago
A 40-lb container of peat moss measures 14× 20× 30 in. A 40-lb container of topsoil has a volume of 1.9 gal. Calculate the avera
Viktor [21]

The average density of the peat moss in units of g/cm³ is 2.52 g/cm³

<h3>How to convert 40 lb to grams (g)</h3>

We'lol begin by converting 40 lb to grams (g). This can be obtained as follow:

1 lb = 453.592 g

Therefore,

40 lb = (40 × 453.592) / 1 lb

40 lb = 18143.68 g

<h3>How to convert 1.9 gal to cm³</h3>

We can convert 1.9 gal to cm³ as follow:

1 gal = 3785.41 cm³

Therefore,

1.9 gal = (1.9 gal × 3785.41 cm³) / 1 gal

1.9 gal = 7192.279 cm³

<h3>How to determine the density </h3>

The density can be obtained as follow:

  • Mass = 18143.68 g
  • Volume = 7192.279 cm³
  • Density =?

Density = mass / volume

Density = 18143.68 / 7192.279

Density = 2.52 g/cm³

Learn more about density:

brainly.com/question/952755

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8 0
2 years ago
I need help with a questino
kipiarov [429]

Explanation:

What will the question be ?

6 0
3 years ago
Read 2 more answers
1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate
Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0
2 years ago
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